Construct ∆ABC in which BC = 4 cm, ∠B = 50° and ∠C = 70°.
Steps of construction:
Draw a line segment BC of length 4 cm.
Draw ∠CBX such that ∠CBX=50°.
Draw ∠BCY with Y on the same side of BC as X such that ∠BCY=70°.
Let CY and BX intersect at A.
ABC is the required triangle.
Draw ∆ABC in which BC = 8 cm, ∠B = 50° and ∠A = 50°.
∠ABC + ∠BCA + ∠CAB = 180°
∠BCA = 180° − ∠CAB − ∠ABC
∠BCA = 180°− 100° = 80°
Steps of construction:
Draw a line segment BC of length 8 cm.
Draw ∠CBX such that ∠CBX = 50°.
Draw ∠BCY with Y on the same side of BC as X such that ∠BCY = 80°.
Let CY and BX intersect at A.
Draw ∆ABC in which ∠Q = 80°, ∠R = 55° and QR = 4.5 cm. Draw the perpendicular bisector of side QR.
Steps of construction:
Draw a line segment QR = 4.5 cm.
Draw ∠RQX = 80° and ∠QRY = 55°.
Let QX and RY intersect at P so that PQR is the required triangle.
With Q as centre and radius more that 2.25 cm, draw arcs on either sides of QR.
With R as centre and radius more than 2.25 cm, draw arcs intersecting the previous arcs at M and N.
Join MN
MN is the required perpendicular bisector of QR.
Construct ∆ABC in which AB = 6.4 cm, ∠A = 45° and ∠B = 60°
Steps of construction:
Draw a line segment AB = 6.4 cm.
Draw ∠BAX = 45°.
Draw ∠ABY with Y on the same side of AB as X such that ∠ABY = 60°.
Let AX and BY intersect at C.
ABC is the required triangle.
Draw ∆ABC in which AC = 6 cm, ∠A = 90° and ∠B = 60°.
∠A + ∠B + ∠C = 180°
Therefore ∠C = 180°− 60°− 90°= 30°
Steps of construction:
Draw a line segment AC = 6 cm.
Draw ∠ACX = 30°.
Draw ∠CAY with Y on the same side of AC as X such that ∠CAY = 90°.
Join CX and AY. Let these intersect at B.
ABC is the required triangle where angle ∠ABC = 60°.